Bateman has observed that for $h\geq 3$ there is a basis of order $h$ with no restricted order, taking \[A=\{1\}\cup \{x>0 : h\mid x\}.\] Kelly [Ke57] has shown that any basis of order $2$ has restricted order at most $4$ and conjectured it always has restricted order at most $3$ (which he proved under the additional assumption that the basis has positive lower density). Kelly's conjecture was disproved by Hennecart [He05], who constructed a basis of order $2$ with restricted order $4$.

The set of squares has order $4$ and restricted order $5$ (see [Pa33]) and the set of triangular numbers has order $3$ and restricted order $3$ (see [Sc54]).

Is it true that if $A\backslash F$ is a basis for all finite sets $F$ then $A$ must have a restricted order? What if they are all bases of the same order?

Hegyvári, Hennecart, and Plagne [HHP07] have shown that for all $k\geq2$ there exists a basis of order $k$ which has restricted order at least \[2^{k-2}+k-1.\]